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            <h1 id="seo-header">『算法-ACM竞赛-真题』2019ICPC南京网络赛FGreedySequence</h1>
            
            
              <div class="markdown-body">
                
                <h1 id="『算法-ACM-竞赛-真题』2019ICPC-南京网络赛-FGreedySequence"><a href="#『算法-ACM-竞赛-真题』2019ICPC-南京网络赛-FGreedySequence" class="headerlink" title="『算法-ACM 竞赛-真题』2019ICPC 南京网络赛 FGreedySequence"></a>『算法-ACM 竞赛-真题』2019ICPC 南京网络赛 FGreedySequence</h1><h1 id="2019-ICPC-南京网络赛-F-Greedy-Sequence"><a href="#2019-ICPC-南京网络赛-F-Greedy-Sequence" class="headerlink" title="2019 ICPC 南京网络赛 F Greedy Sequence"></a>2019 ICPC 南京网络赛 F Greedy Sequence</h1><p>You’re given a permutation aa of length nn (1 \le n \le 10^51≤n≤105).</p>
<p>For each i \in [1,n]i∈[1,n], construct a sequence s_isi​ by the following rules:</p>
<ol>
<li>s_i[1]&#x3D;isi​[1]&#x3D;i;</li>
<li>The length of s_isi​ is nn, and for each j \in [2, n]j∈[2,n], s_i[j] \le s_i[j-1]si​[j]≤si​[j−1];</li>
<li>First, we must choose all the possible elements of s_isi​ from permutation aa. If the index of s_i[j]si​[j] in permutation aa is pos[j]pos[j], for each j \ge 2j≥2, |pos[j]-pos[j-1]|\le k∣pos[j]−pos[j−1]∣≤k (1 \le k \le 10^51≤k≤105). And for each s_isi​, every element of s_isi​ must occur in aa <strong>at most once</strong>.</li>
<li>After we choose all possible elements for s_isi​, if the length of s_isi​ is smaller than nn, the value of <strong>every undetermined element</strong> of s_isi​ is 00;</li>
<li>For each s_isi​, we must make its weight high enough.</li>
</ol>
<p>Consider two sequences C &#x3D; [c_1, c_2, … c_n]C&#x3D;[c1​,c2​,…cn​] and D&#x3D;[d_1, d_2, …, d_n]D&#x3D;[d1​,d2​,…,dn​], we say the weight of CC is <strong>higher than</strong>that of DD if and only if there exists an integer kk such that 1 \le k \le n1≤k≤n, c_i&#x3D;d_ici​&#x3D;di​ for all 1 \le i &lt; k1≤i&lt;k, and c_k &gt; d_kck​&gt;dk​.</p>
<p>If for each i \in [1,n]i∈[1,n], c_i&#x3D;d_ici​&#x3D;di​, the weight of CC is equal to the weight of DD.</p>
<p>For each i \in [1,n]i∈[1,n], print the number of non-zero elements of s_isi​ separated by a space.</p>
<p>It’s guaranteed that there is only one possible answer.</p>
<h4 id="Input"><a href="#Input" class="headerlink" title="Input"></a>Input</h4><p>There are multiple test cases.</p>
<p>The first line contains one integer T(1 \le T \le 20)T(1≤T≤20), denoting the number of test cases.</p>
<p>Each test case contains two lines, the first line contains two integers nn and kk (1 \le n,k \le 10^51≤n,k≤105), the second line contains nn distinct integers a_1, a_2, …, a_na1​,a2​,…,an​ (1 \le a_i \le n1≤ai​≤n) separated by a space, which is the permutation aa.</p>
<h4 id="Output"><a href="#Output" class="headerlink" title="Output"></a>Output</h4><p>For each test case, print one line consists of nn integers |s_1|, |s_2|, …, |s_n|∣s1​∣,∣s2​∣,…,∣sn​∣ separated by a space.</p>
<p>|s_i|∣si​∣ is the number of non-zero elements of sequence s_isi​.</p>
<p>There is no space at the end of the line.</p>
<p>样例输入复制</p>
<pre><code class="hljs">2
3 1
3 2 1
7 2
3 1 4 6 2 5 7
</code></pre>
<p>样例输出复制</p>
<pre><code class="hljs">1 2 3
1 1 2 3 2 3 3
</code></pre>
<p>这题是队友出的，这个题好像线段树能做，但是我们是暴力加贪心做的，前几遍是枚举的区间位置纯暴力，后来是枚举左右区间长度，对于每个元素的下一个元素都是唯一固定的。之后记忆化搜索就可以求出每个序列。</p>
<pre><code class="hljs">#include&lt;stdio.h&gt;
#include&lt;algorithm&gt;
#include&lt;iostream&gt;
#include&lt;string.h&gt;
#include&lt;math.h&gt;
#include&lt;vector&gt;
#include&lt;queue&gt;
#include&lt;map&gt;
using namespace std;
#define LL long long
#define MAXN 100100
int a[MAXN],pos[MAXN];
int s[MAXN],oo[MAXN];
int main()
&#123;
    int t;
    scanf(&quot;%d&quot;,&amp;t);
    int n,k,l,r,maxx;
    while(t--)&#123;
        scanf(&quot;%d%d&quot;,&amp;n,&amp;k);
    for(int i=1;i&lt;=n;i++)&#123;
        scanf(&quot;%d&quot;,a+i);

        pos[a[i]]=i;
        s[i]=1;
    &#125;


    s[0]=0;
    for(int i=2;i&lt;=n;i++)&#123;
        int j=i-1;
        while(j&gt;0)&#123;
            if(abs(pos[i]-pos[j])&lt;=k)&#123;
                s[i]+=s[j];
                break;
            &#125;
            j--;
        &#125;
    &#125;
    for(int i=1;i&lt;=n;i++)&#123;printf(&quot;%d&quot;,s[i]);if(i&lt;n)printf(&quot; &quot;);&#125;

    printf(&quot;\n&quot;);
    &#125;
&#125;
</code></pre>

                
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